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If 4 < (7 - x)/3, which of the following must be true?

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chetan2u

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12 Aug 2021, 00:37

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NareshGargMBA

Since the range presented in the Option [II] is x<-5 OR x>-1, how can the expression x<-5 be a MUST BE TRUE condition?
Shouldn't it make more sense with a Could be true statement?

Whenever x<-5 or x>-1, |x+3|>2 is true.

We know from the main statement x<-5, so option II must be true. Of course whenever x>-1, it will still be true.

A simpler way to look at it.
Say statement says x=-2

Option two says x^2=4…x=2 or x=-2.

Now x can be 2 too, but we are concerned with what is the case when x=-2.
Thus x=-2 means |x+3|>2 is always true.

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07 Sep 2022, 04:51

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From stem x<-5 so x=-6, -7 ,-8….

From statement 2 we have
1) x<-5 (which goes with the stem, so for this part of the inequality it is true)
2) x>-1, so x= 0 1 2…
Since the 2nd part does not go with the stem, then how it is accepted as a correct choice?

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07 Sep 2022, 04:57

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NareshGargMBA

Bunuel

SOLUTION

If 4<(7-x)/3, which of the following must be true?

I. 5<x
II. |x+3|>2
III. -(x+5) is positive

(A) II only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II and III

Note that we are asked to determine which MUST be true, not could be true.

$4<\frac{7-x}{3}$ --> $12<7-x$ --> $x<-5$. So we know that $x<-5$, it's given as a fact. Now, taking this info we should find out which of the following inequalities will be true OR which of the following inequalities will be true for the range $x<-5$.

Basically the question asks: if $x<-5$ which of the following is true?

I. $5<x$ --> not true as $x<-5$.

II. $|x+3|>2$, this inequality holds true for 2 cases, (for 2 ranges): 1. when $x+3>2$, so when $x>-1$ or 2. when $-x-3>2$, so when $x<-5$. We are given that second range is true ($x<-5$), so this inequality holds true.

Or another way: ANY $x$ from the range $x<-5$ (-5.1, -6, -7, ...) will make $|x+3|>2$ true, so as $x<-5$, then $|x+3|>2$ is always true.

III. $-(x+5)>0$ --> $x<-5$ --> true.

Answer: D.

Since the range presented in the Option [II] is x<-5 OR x>-1, how can the expression x<-5 be a MUST BE TRUE condition?
Shouldn't it make more sense with a Could be true statement?

I think your doubt is addressed several times in this thread. Have you read it? To understand the underline concept better practice other Trickiest Inequality Questions Type: Confusing Ranges.

Hope it helps.

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19 Sep 2022, 23:05

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For anyone confused with 'Must Be True' questions, consider it this way.
The given range should be a subset of the the range derived from the correct option. Not the other way around.

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21 Mar 2024, 00:37

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This question is of the format - If condition, then what MUST be true. In Quant, such questions can be asked in the context of Number Properties or Algebra concepts. In this question, the context is Algebra.

To answer this question comfortably, one needs to completely understand the condition presented. And this understanding involves visualizing the value of x. And then one needs to carefully simplify the choices. While each individual choice can be visualized on the number line, the reason why many students falter in this question is because they are unable to devise a solid approach to answering this question.

As you watch the video, observe how we have used the process skill of translating Quant back to English to develop an iron-clad approach to solve this question. In fact, statement II is where students tend to falter and by applying this translation technique, they can avoid this pitfall.

Did you also falter on Statement II when you solved this question on your own? Try applying the translation technique going forward!

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01 May 2024, 22:44

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Bunuel

If $4<\frac{7-x}{3}$, which of the following must be true?

I. $5<x$
II. $|x+3|>2$
III. $-(x+5)$ is positive

(A) II only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II and III

Writing down how I solved it-

The qs gives me x<-5

Now what I have to do is identify which of the following is true given that x<-5

1. x>5- I know x<-5 as per the qs. So this is not possible.

2. |x+3|>2
This will be true in 2 cases
a. x+3>2: x>-1
b. -(x+3)>2
i.e -x-3>2: -5>x or x<-5

I know that the range of x<-5. Thus the qs satisfies case 2 of option b. Since it satisfies, I can say that the given question prompt can be moulded into |x+3|>2. And hence it must be true that |x+3|>2 given 4<(7−x)/3

3. -(x+5) is positive
Or i can say that (x+5) is negative. i.e x+5<0
Thus x<-5.
Hence this must be true.

Thus D

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05 Sep 2025, 02:12

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This is a classic GMAT trap that tests your ability to work systematically with inequalities and Roman numeral statements. Many students rush into evaluating the statements without first establishing the constraint on x.

Strategic Framework:

Step 1: Solve the Core Constraint
Given: (\frac{4-7x}{3} > 3)
Multiply both sides by 3: (4-7x > 9)
Subtract 4: (-7x > 5)
Divide by -7 (flip the inequality): (x < -\frac{5}{7})
Wait - this means (x < -5), not (x > -5). Critical insight: The constraint severely limits our x values.

Step 2: Systematic Statement Evaluation
Now that we know (x < -5), let's check what MUST always be true:
Statement I: (x > 5)
Since (x < -5), this can never be true. FALSE

Statement II: (|x + 3| > 2)
If (x < -5), then (x + 3 < -2)
Since (x + 3) is negative and less than -2: (|x + 3| = -(x + 3) > 2) ✓ TRUE

Statement III: (-(x + 5)) is positive
Since (x < -5), we have (x + 5 < 0)
Therefore (-(x + 5) > 0) ✓ TRUE

Answer: D) II and III only

The key insight here is recognizing that "must be true" problems require you to find what's always true given the constraint, not what's sometimes true. This pattern appears frequently in GMAT inequalities.

For the complete breakdown showing the systematic approach to all Roman numeral inequality problems, plus the 3 most common trap patterns students fall into: https://neuron.e-gmat.com/quant/questions/if-4-7-x-3-which-of-the-following-must-be-true-1617.html

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05 Sep 2025, 10:19

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Hope it will be helpfull for everyone

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Hope it will be best

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28 Feb 2026, 14:36

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Bunuel

If $4<\frac{7-x}{3}$, which of the following must be true?

I. $5<x$
II. $|x+3|>2$
III. $-(x+5)$ is positive

(A) II only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II and III

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